The correct option is C [−12,1]
x2+y2+z2=1 …[1]
(x−y)2+(y−z)2+(z−x)2=2(x2+y2+z2−xy−yz−zx)
(x−y)2+(y−z)2+(z−x)2=2[1−(xy+yz+zx)]
xy+yz+zx=1−12[(x−y)2+(y−z)2+(z−x)2]
xy+yz+zx≤1 …[2]
As [(x−y)2+(y−z)2+(z−x)2≥0]
Again,
(x+y+z)2=x2+y2+z2+2(xy+yz+zx)
⇒(x+y+z)2=1+2(xy+yz+zx)
⇒xy+yz+zx=12[(x+y+z)2−1]
⇒xy+yz+zx≥−12 …[3]
[∵(x+y+z)2≥0]
From [2] and [3],
xy+yz+zx∈[−12,1]